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March 9, 2018 05:20
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srli revised this gist
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This file contains hidden or bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters. Learn more about bidirectional Unicode charactersOriginal file line number Diff line number Diff line change @@ -68,7 +68,7 @@ def solve(self): def answer(maze): astar = AStar(maze) return astar.solve() maze = [[0, 0, 0, 0, 0, 0], [1, 1, 1, 1, 1, 0], -
srli created this gist
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This file contains hidden or bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters. Learn more about bidirectional Unicode charactersOriginal file line number Diff line number Diff line change @@ -0,0 +1,80 @@ #!/usr/bin/python import heapq class Cell(object): def __init__(self, x, y, is_wall): self.reachable = not is_wall self.x = x self.y = y self.parent = None # G, H, F are current costs self.g = 0 # cost to get to end pos self.h = 0 # cost to get to from start pos self.f = 0 # g + h class AStar(object): def __init__(self, maze): # Open cells to still visit self.opened = [] heapq.heapify(self.opened) # Visited cells list self.closed = set() # Initialize all cells self.init_grid() def init_grid(self): # Create cells by looping through the input maze for x in range(self.grid_width): for y in range(self.grid_height): if self.maze[y][x] == 1: is_wall = True else: is_wall = False self.cells.append(Cell(x, y, is_wall)) def solve(self): # Add starting cell to open heap queue heapq.heappush(self.opened, (self.start.f, self.start)) while len(self.opened): # Pop cell from heap queue f, cell = heapq.heappop(self.opened) # Add cell to closed list so we don't process it twice self.closed.add(cell) # If ending cell, return found path if cell is self.end: return self.get_path() # Get adjacent cells for the current cell adj_cells = self.get_adjacent_cells(cell) for adj_cell in adj_cells: # If the adj_cell is not a wall and hasn't been visited if adj_cell.reachable and adj_cell not in self.closed: # If adj_cell in open list from another cell, check if current path is # better than the one previously found for this cell if (adj_cell.f, adj_cell) in self.opened: if adj_cell.g > cell.g + 10: # If so, update its cost and parent with the current cell self.update_cell(adj_cell, cell) else: # If adj_cell has never been visited before, update it with its # cost and parent cell. Add to heap for use self.update_cell(adj_cell, cell) heapq.heappush(self.opened, (adj_cell.f, adj_cell)) def answer(maze): astar = AStar(maze) return astar.solve_all_paths() maze = [[0, 0, 0, 0, 0, 0], [1, 1, 1, 1, 1, 0], [0, 0, 0, 0, 0, 0], [0, 1, 1, 1, 1, 1], [0, 1, 1, 1, 1, 1], [0, 0, 0, 0, 0, 0]] print answer(maze)